Question

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If it takes 17.5 mL of 0.085 M NaOH to titrate a 15 mL sample of sauerkraut juice, what is the acidity of that juice, expressed as % lactic acid (wt/vol)?

Answer 1

First we find moles of NaoH used

Moles = M x V ( in liters)    , whre vol of NaOH used = 17.5 ml = 0.0175 L

Moles of NaOH = ( 0.085 x 0.0175) = 0.0014875

Acid and base reacts in 1:1    

( Let HA be lactic acid representation , then reaction is HA + NaOH --> NaA + H2O)

Hencee lactic acid moles at end point = NaOH moles used = 0.0014875

Lactic acid mass = Moles x molar mass of lactic acid

            = 0.0014875 mol x 90.08 g/mol

          = 0.134 g

Now 0.134 g lactic acid is present in 15 ml

lactic acid % ( w/v) = ( 100 x lactic acid mass in g / solution volume in ml)

                      = ( 100 x 0.134 / 15)

                  = 0.893 %